Video lecture: oxidation reduction
The video presents a lecture on oxidation-reduction. Oxidation-reduction is important for many biological processes. Edited by Ashraf
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Added: April 16, 2009, 11:26 pm
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Chemical Science - The Shapes of Molecules: VSEPR Theory
Principles of Chemical Science/nVideo Lectures - Lecture 29/nTopics covered: /nThe Shapes of Molecules: VSEPR Theory/nInstructor: /nProf. Catherine Drennan/nTranscript - Lecture 29/nMonday's class really picks up from the lecture we had this past Monday. We started crystal field theory and then talked about octahedral-shaped molecules and splitting of d-orbitals. And then on Monday, we are going to continue with that topic./nWe are going to move in and talk about tetrahedral molecules, square planar molecules and how they are going to split the d-orbitals. And then how that relates to something, why we need to know about any of that. And that is because it is able to predict some of the properties of molecules, like their colors and also their properties in terms of spin and using various different spectroscopies to learn something about particular molecules, coordination complexes./nBefore we get into that unit, we are going to kind of take a step back and talk today about shapes of molecules. We introduced octahedral, but now we are going to talk about all the other kinds of shapes of molecules in today's lecture. This is kind of jumping around in the book a little bit. We have been in Chapter 16. And now, the way the book is divided, this particular unit is back in Chapter 3, although it is referred to a lot in Chapter 16. This is kind of some of the things that you need to know to finish up Chapter 16. And then on Monday we will be back in Chapter 16 as well. Again, this is a little bit odd order because of the way the calendar fell this year. All right. VSEPR, shapes of molecules, why do we need to know about this?/nWell, as most of you know, I do protein crystallography, which is all about structure. So, of course, I am going to think that structure is important, that shape and geometry are important. And, in fact, they have an influence on the physical properties and chemical properties of molecules. Things like melting points, boiling points, reactivities, and, in particular, in biological systems it is important./nA biological system has particular shapes of chiral molecules, it is a chiral environment, which we have talked about before. And so molecules must fit precisely in that environment. Often, if you are talking about an inhibitor of an enzyme, the inhibitor or the pharmaceutical drug is designed so it specifically fits into a particular crevice in an active site and blocks activity. A lot of the world of pharmaceuticals is all about shape of molecules and how you design particular types of shapes of molecules./nWe have learned a lot about this sort of area already, and we found that the shapes of molecules can be predicted pretty well using simple methods based on Lewis structures and also in terms of thinking about sort of where orbitals are and how that may affect the particular shape of the molecules. Again, this is sort of a simplistic view that works pretty well. And it will be also, on the next problem set, a good opportunity for you to review Lewis structures before you take the final. Sometimes this unit is with Lewis structures, there is some rationale for that, but it is also kind of nice to come back to it later in the semester as a way to review things that you learned before and get you ready for the final exam. The particular theory we are going to talk about today is referred to as VSEPR. And this is the valence-shell electron-pair repulsion theory./nAnd so, like crystal field theory, you are talking about repulsive effects. Here this is based on a simple principle that valence electron pairs will repel each other and that the geometry around a central atom will be such to minimize that repulsion. This is a lot like crystal field theory, where we were talking about the splitting of the d-orbitals. And we are talking about ligands as negative point charges heading toward those d-orbitals. A lot of these theories that work pretty well are just really based on the idea that repulsion is unfavorable. It is a pretty simple concept which can explain a lot in terms of properties of molecules. Before we get into this, there is always nomenclature that we must deal with. And here is the nomenclature here. We have A as a central atom./nWhatever is in the middle, which when we are talking about transition metals, complexes would be your metal, but this works for organic chemistry as well, when you are all talking about carbon, oxygen, sulfur, nitrogen. And X is going to be your bound atom, and E will be a lone pair of electrons, hence the E. There are just a few rules. It is not too bad compared to some of the other things we have done. Just a handful of rules that you need to know. First, you need to know that the steric number, abbreviated S.N., is used to predict geometries, and what steric number refers to. It refers to the number of atoms bound to central atom, plus the number of lone pair electrons on the central atom./nHere we are not concerned about double or triple bonds. We are only concerned about how many things are around our central atom, both the bonded and the lone pairs. Whether it is a double or a triple bond doesn't matter at all for these cases. It is the number of atoms bonded, not the number of bonds, which is important to remember. And so steric number, here, should not be confused with coordination number, which we talked about in transition metals, --/n-- which is just the number of ligands around the central metal. So, there are a number of different things that you are going to learn about. All right. For resonance structures, you can imply a VSEPR to any one of them. And there are two cases that we need to consider today. One is the case when there are no lone pairs, and other is the case when there are lone pairs. Those are all the rules. It is really not too complicated for this. We are just going to go through today and look at all these different molecules and consider the nomenclature and the geometry around them. First, we will consider formula type./nAnd so AX two would mean that you have a central atom A, and we have two atoms bound to it, so two Xs. And now we are considering the cases without lone pairs of electrons, so there won't be any Es until later. Then the steric number is going to be two. And so here would be the shape of the molecule. And so we have a center atom and two atoms bound to it. It would have linear geometry and a bond angle of 180./nThat is pretty simple. Now, we go to the next category, where you have three things bound, so that would have an SN number of three. And this is going to be the shape of that molecule. And what is this called? Trigonal planar. And what are the angles? 120. Okay. Now, if we move up to four, and this one should be pretty familiar to you, if we have four things bound, SN number of four, this is the molecular shape that we have. And you all know this as tetrahedral. And the angles of tetrahedral are 109.5. These are all things you have to memorize./nThis is written in your notes, but if you don't look at your notes and just try to think about it, it will be a good review for the next exam on this topic. All right. If we go into the next category, when we have five things bound, that would be AX five, five atoms bound, SN number of five. We have this trigonal bipyramidal geometry. And what are the angles around the equator here connecting the red atoms? 120. And what about red to black? 90. All right. Then, if we go to the next category, when we have six things, we talked a lot about this in the last two classes, what kind of geometry is this? It is octahedral geometry, with angles of 90./nRight. These are all the basic shapes when you just have two through six atoms bound. Let's take a look at a couple of examples of molecules when we have actual atoms listed. If we look at CO two, what is our SN number? Two, right. We are only considering the number of atoms bound. We don't care that they are double bonds. We are only concerned with the atoms that are linked to the central atom here. Then the geometry is what? Linear. And angles? 180, right. We will look at another example here. Is this one going to be the same or different from the one above? The same, right, because we don't care about the double or triple bonds. We can go through. SN number is still two./nIt is still linear. And the angle is still 180. We move into another example here. And I will just say that these lecture handouts have this information in here. This is the most requested lecture handout that I have from people who have been in the classes who get up to other level classes and need to know this stuff. And, for some reason, some of them have not kept their 5.111 notes sort of next to their pillow in later years./nI am not really sure why, but sometimes it gets lost and they request this handout, because it is a really nice summary of all of these things. You might want to keep it in a nice special place for future reference. All right. Some day I should bring in what my freshman chemistry textbook looked like with like everything highlighted and stuff just for entertainment value. All right. This molecule here would have a SN number of three. And its shape again? Trigonal planar. And angles? 180. By the time you get through with lecture today, you should have all of this memorized, because we are going to see these a couple more times. All right. This one, what molecule is this? Does someone know its name? Methane, right. Again, what is this geometry?/nTetrahedral. SN number of four. And angles of? 109.5, right. Okay. Now we get into a few more complicated molecules that you do not see quite as often, although I think we have seen this one quite often in different kinds of problems, including chemical equilibrium and other things. SN number here? Five. Geometry? It is hard to say that one in unison. We can work on this for the end of lecture. Then we can all say it in harmony. All right. And our angles again? Yeah. Okay. And one last one. This should be very familiar to you. This is SN number six. Geometry? Octahedral. Angles? 90. Okay. People are doing well./nHere is some stuff that you may not have heard before, or maybe have in another chemistry course, but now we are going to introduce lone pairs and see what effect this has. It gets a little bit more complicated at this point. For lone pairs, we think of them as being sort of centrally located. And I brought in some props to talk about the spatial distribution of electrons. One can imagine that the electrons are spread out as far as they can go. They like to have a lot of space. And so the lone pair electrons are going to take up more space and experience more repulsion then when you have the electrons as part of bonds to atoms. The lone pairs are not sort of held in as much, they are not sort of forced to be in one location, so they will spread out more./nAnd so because of this, their spreading out more, they are taking up room, they are kind of pushing everything away, because they want to have as much room as they possibly can. That means that lone pairs are going to be more repulsive than the atoms that are bound. Even though electrons are involved in those bonds, they are not quite as bad. The worst thing you can have is two sets of lone pairs near each other. They want all the room./nAnd so they are going to be sort of bumping into each other and trying to force the other set farther away. See, this is much more bulky than if you just have the two atoms bound with bonds to a central atom. The next would be a lone pair versus a bonding pair. And then the least repulsive will be two bonding pairs because those electrons are sort of fixed into locations for the bonds. They are not requiring as much space./nThey are not sort of spreading out and taking over all the space that is available. Worst is lone pair-lone pair, then lone pair bonding, and then bonding-bonding. And, if you think about this sort of simple rule, you can predict a lot about the shapes of molecules. Let's look at what shape something will have if it has this sort of set molecule. Again, A, the central atom. X is four atoms that are bonded to the central atom, and E is one set of lone pair. And so this is going to have what is known as a seesaw shape. And let's think about why that would be the case. If we have this sort of central framework, but one of these bonded atoms is now going to be replaced with a lone pair. There are five things around this framework./nOne is a lone pair. Where are you going to put it? If we put it on top, what are the angles between these lone pairs on top and the three red bonded atoms? 90. That is kind of close and not so favorable. If we instead put one over here, then we have still the sort of negative interaction with the 90 here and the 90 here. But we only have two of them instead of three. And these guys are 120 apart, so it is not so bad. There is a little bit more space in the equatorial position. And so that is why you get this seesaw shape, because you would want to put it in the equatorial position, where there is only sort of these two rather than three./nThere is just a little bit more space, because these guys are farther away. That gives rise to seesaw geometry. You can sort of again, these rules are more to explain in retrospect what is observed about those molecules, and then you rationalize. But the rationalization works pretty well. If we do that, if we put a lone pair, then, here, instead of one of these top positions, we get this seesaw geometry./nAnd so let me help you remember this. See? Get the seesaw geometry. How many of you actually grew up with seesaws in the neighborhood playground? Still most people, okay. They are actually one of the most dangerous toys. And I think a lot of playgrounds have removed them because of the one kid who sort of sits on it and keeps you, like, stuck up in the air because they are a little heavier than you are./nAnd you are stuck there for a really long time, and then eventually they get up and you go falling and flying. We may have to give this a new name at some point when all seesaws are removed from playgrounds, but it sounds like it is still usable. Now, if we wanted to put another lone pair on that same structure, we still have essential atom, three atoms bound. Now we have two lone pairs. Those two lone pairs, then, will go into the equatorial positions as well. And so you kind of have this sort of arrangement. And they were so repulsive they actually made the atom unbonded from it. Okay./nThat leaves you with this shape, which is a T-shape molecule. Again, this is a rationalization. Someone figured out that a molecule like this had this T-shape ad it could be rationalized by the fact that the lone pairs wanted to sort of fill out in that equatorial plane. And they could move away slightly from each other, and that would minimize the amounts of repulsion. All right. If we keep going here and look, now, at a molecule with this geometry, AX four E two molecule. That is based on kind of what structural framework? What is the sort of parent structure when you have six things bound? Yeah, based on an octahedral geometry./nIf you take your octahedral molecule and if you want to figure out where you are going to put your lone pairs, you could stick one down here in the equatorial, but that doesn't do you as much good this time because now your other equatorial things are at 90. This one would have a lot of crashes down here. What ends up happening is the two lone pairs would go one above and one below, so they would be as far away from each other as possible. You have one below and another one on top. And remember, the worst thing to have is a lone pair-lone pair. Even though you have four atoms at 90 that it is interacting with, at least it is far away from the other lone pair./nEach kind of gets half the molecule, one up top and one on the bottom, so they have the most room, those electrons, to spread out. And so that gives you what kind of geometry? Square planar, right. All right. You can also predict something about what might happen to the angles that you have all now pretty much memorized on the previous slides. That if you do have a lone pair, and out here we have an AX three E, SN number of four, so four things bound, one lone pair and three atoms, and this is based then on what kind of structure? With SN number of four, what is the kind of the parent geometry there? Yeah, tetrahedral is the parent geometry./nAnd now we have replaced one of the bonded atoms with a lone pair. And so what that is going to do is that lone pair is really not in that bad of shape. You put it on top. You have 109.5 now between those angles pointing down in the parent geometry. And that was better than the others which were at 90, but it is still not good enough for it. And so it actually presses down even more. Those lone pairs, they are going to spread out as much as they possibly can and repel those bonded atoms away to the extent possible. In this molecule, what is observed is that the angles in between here, in between the hydrogens, instead of being 109.5, which is what you would expect for the true tetrahedral geometry, they are 106.7. This lone pair is pressing down, compressing the bonded atoms even closer to each other than what is sort of the normal framework./nThis really takes up a lot of room here and is repulsive toward these bonded angles. Again, the VSEPR theory can predict these kinds of, or really explain these kinds of changes. If you observe the smaller angle you can say, oh, there was a lone pair. That makes sense that it would be a smaller angle because there would be repulsion down. Again, you can think about it. You can make one of these lone pairs yourself, low tech and think about the shapes here. All right. We can also rationalize the difference between a PH3 and NH3 molecule. They have pretty much the same framework here, based on tetrahedral geometry AX3E, SN number of four./nBut now, instead of the number for methane that we saw of 109.5, and instead of the 106.7 for NH3, we now get a really small angle between here, between the angle between hydrogen and phosphorus. And hydrogen is now 93.3, so it is even more compressed. And the reason why is because we are moving down in the Periodic Table. And so as we move down the column in the Periodic Table, we are going to have lone pairs that have even larger spatial volumes. And so they are going to compress even more. Now it is an even bigger sort of situation here. The lone pairs are trying to fill up even more volume./nAnd so they are sort of trying to compress as much as possible these atoms away. They are filling up all available space and making the angle here even smaller. And so if asked about this, if asked to predict, you could sort of look at where things are in the Periodic Table. Here is nitrogen, here is phosphorus. And so if we had a compound based on nitrogen or phosphorus that was based on tetrahedral geometry, one might expect that the angle becomes even smaller as you move down the Periodic Table, as that lone pair that is on there is filling up even more and more spatial volume. This kind of combines ideas of trends in the Periodic Table with shapes of molecules. All right. Another thing that you can sometimes predict is about bond lengths./nIf you experimentally observed different bond lengths in a molecule, you could use VSEPR to go back and think about where that may have come from. What is this geometry here, the one that is shown and also the parent geometry? The one that is shown is what? Seesaw. And the parent geometry? Right, trigonal bipyramidal. All right. Here we have AX four E. We have an SN number of five. We have four atoms bound. We have one lone pair. And now it is observed that there are two sets of SF bonds. One set is 1.545, and one is 1.646. What do you think those different bond sets --/nThere are two types of bonds. We have one lone pair. Likely one set would involve the gray atoms, and the other would be the black atoms. Which do you think would be longer? Do you think the gray or the black would be longer, the bond lengths from the central atom to the gray or to the black would be longer? Gray. And why is that? Right, because these are at 90 going down. Whereas, this is at 120. The 90 is more repulsive, so it basically kind of pushes that up farther and down farther, moves them farther away. The black atoms are already farther and won't feel the repulsion as much./nAnother way to sort of, if you have a lone pair here, for the lone pair to feel like it really has the space that it requires, is to sort of repel the atoms that are closer in terms of distance, repel them just a little bit farther away. And then it can kind of fill out and feel like it has enough space. Again, if you had an experimentally observed set of bond lengths, you could rationalize using VSEPR, why you would have this./nOkay. As we said, more repulsion between the lone pair electrons and the axial ones because those are at 90, so those would move farther away. That is a lot of the theory. Now we are just going to run through these sheets and talk about the different types of geometries, the names of the different types of geometries when you have the lone pairs on there. And then look at some examples. Again, if you call out the answers, and if we go through this quickly, this is all that we need to go over for today, and it will be a good review. You will sort of have the opportunity, today, to memorize this. You might want to go back over it a couple times, but pretty much this is the whole unit. It is not too challenging. If we have AX two E, we are going to have a SN number of three./nAnd a molecular shape. I will draw that. And what is the name of that geometry? It is bent. And that is because you are only considering the bonded atoms. You are not talking about the lone pair. We don't really see the lone pair, two bends, and so it would just look bent to us. That is where that comes from. And what angles would this have for this particular case, SN number of three? Okay. We have this one here. The SN three would have a trigonal planar parent geometry./nBut if we take a lone pair and put that on instead of an atom, then it will also be somewhat compressing of those two atoms down there. We won't really know, without knowing what the central atom is, we cannot really predict how bad the compression is going to be. What you would put is less than 120 here, because it would be 120 if this was an atom and not a lone pair. If this is a lone pair, then it is less than 120./nAll right. Here is another case. Now we have three atoms bound, one lone pair, SN number of four. And our parent geometry here would be tetrahedral. And now we have a lone pair. We saw this a little bit before. And what is that geometry called? Here, we have this picture. What is this called? Trigonal pyramidal. We have this. And what would the angles be here? Less than 109.5. All of you should know the answers because they are actually in your handout, so feel free and confident to yell them out. All right. Again, this is going to be less than 109.5 because we have a lone pair pressing down./nYou have to in all these cases consider what the parent geometry is. And then the angles are going to be a little less than that if you have one lone pair sitting on top. All right. AX two E two has a SN number of four. All right. What geometry is this based on? Yeah, tetrahedral again. We have two lone pairs on that. And what does that look like it should be called? Bent as well. And now the angles are going to be less than 109.5. So, you see, we have two bent molecules, but the angle isn't the same. It depends on whether it is based on a trigonal planar or it is based on tetrahedral. You need to sort of know what the whole molecule was to figure out what degree of bend you have in these two./nAll right. AX four E has SN number of five. And so our shape here, what is this one going to be? Yeah, this is going to be a seesaw. And so what are our angles going to be? Right, less than 120, less than 90. It could be that really the 120 are not going to be that affected, but that will be what you will put down because you don't really know. This is just a rough predictor of what those are going to be, so it is less than what the parent molecule was. All right. SN number of five. And so what are we going to be doing here? We are going to add another lone, take off an atom, add another lone pair./nAnd what are we going to get here? Right, T-shaped. And the angles will be... At this point is our T-shape, and so we will have less than 90. You will have the two lone pairs in the equatorial position, as we talked about before, trying to spread out as much as they possibly can. Now we are replacing another one with a lone pair. Another bonded atom. Our SN number is still five. And what shape are we going to get in that case? Linear, right. They all go in these equatorial positions to spread out as much as they possibly can. And so we are going to end up with a linear molecule, so we would replace here. And what are our angles here? They are exactly 180. Yeah?/nIt could be that there would, but they probably will just be really spreading out around to fill out as much room as possible around the whole equatorial place. You can just sort of think about it as if there was space, they would move into it, so they have really filled up this space. And pretty much you might as well keep them there, because whichever way they move, they are just going to get repelled in another direction./nThat's kind of the one exception, are these ones that turn out to be linear. There is one other exception where the angles are not going to change as well, where you have orbitals on kind of either side that counteract. That is a good question. All right. One more. We have AX five E, SN number of six. This is based on what parent molecule? Octahedral. We are going to take off one ligand, and so this would be, if we removed one of the ligands here, we get this parent shape, which is called square pyramidal. And what are the angles going to be, here? Less than 90. Okay./nEvery year, when I do this unit, I like to reward people who are in class. And so I give a little hint about the final. And the hint will be that this will be on the final, [LAUGHTER] so you might want to put that down. The last couple of years I did seesaw, which really is my personal favorite, but at some point I have to switch. This time it will be this one. And it is always so sad to me when people get that one wrong, so please get that one right. Okay. Let's keep going. SN number is six. And so we are going to put on another lone pair. And where are we going to put it? We are going to try to move it so that there is the least amount of repulsion between the lone pairs, so it is going to be up on top./nThis geometry keeps the lone pairs as far away from each other as possible. And what is the remaining thing called when you have those lone pairs on there? That is going to be square planar. And what about the angles in this case? In this case they are just 90, because you have a lone pair on top pushing down and a lone pair on the bottom pushing up. This is another example where it is not less than. All right. Next one. We add another lone pair. And at this point does it matter where you put the lone pair? No. They are all equivalent positions, so it doesn't make any difference. It is going to be one of these guys gets replaced, and so that is going to leave you with the what geometry?/nT-shaped. And angle here is going to be this time less than 90, right, because it will repel this one and this one if you have a big lone pair in this position. All right. One more. Again, the same geometry. We take another bonded atom off and put another lone pair. And what are we left with in this case? Linear. Again, this is going to be 180. All right. Now we have looked at all of those, and we can very quickly go down. And in this handout, I have most of it written. I don't have the angles, so that you can write the angles in and have your complete set for this as well. Now let's look at water. The Lewis structure for water, you would end up with two lone pairs on the oxygen./nAnd so what formula type would this be? AX two E two. SN number of four. Geometry? Bent angle. Less than 109.5, because this is based on the tetrahedral geometry. Remember, bent could either be less than 120 or less than 109.5, but if you have a SN number of four then it is based on tetrahedral geometry. It would be less than 109.5, which is the tetrahedral angles. Okay. SF four, here is our structure. Formula type for this would be? AX four E. And that is a SN number of five. And the shape? Seesaw./nAnd angles? Yeah, less than 90 and less than 120. Here is another one, BrF three. These are all probably ones that you did Lewis structure for. If not, I am sure you will before the final. Here is another one. The formula type for this would be AX three E two. And SN number? Five. And geometry? T-shaped. Angles? Less than 90. Okay. If we keep going, here is another example, selenium and fluoride. And we have three sets of lone pairs. What is the formula type? AX two E three. SN number of five. Geometry? Linear. Angle?/n180. Okay. BrF five. Formula type? AX five E. SN number? Six. Geometry? Square pyramidal. Angles? Less than 90. You want to remember that one. Okay. Here is another one with an addition lone pair. Formula type? AX four E two. SN number? Six. Geometry? Square planar. Angle? 90, exactly. That completes that list. Again, the theory of VSEPR, you are taking experimental results and then trying to apply a theory to it. It is not good at really making exact predictions about things./nBut it is really good at rationalizing what they have. All right. Have a wonderful Thanksgiving, everybody.
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Added: April 16, 2009, 11:33 pm
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Chemical Science - Kinetics - Lecture 31
Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 31/nTopics covered: /nKinetics/nInstructor: /nProf. Catherine Drennan
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Added: April 16, 2009, 11:37 pm
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Chemical Science -Transition Metals - Lecture 27
Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 27/nTopics covered: /nTransition Metals/nInstructor: /nProf. Catherine Drennan/nTranscript - Lecture 27/nWe finished the unit on oxidation-reduction on Wednesday./nAnd so now we are moving into the next unit, which is transition metals, and that is Chapter 16 in your book. And following transition metals, the last unit is on kinetics or rates of reaction, and that is a little bit of a longer unit./nSo there are only two more units coming up in the course. Today I am going to introduce you to the transition metal unit, and this follows nicely from oxidation-reduction since you saw a lot of electrons being lost and gained./nAnd metals are definitely something that will gain and lose electrons, so it follows nicely from that./nToday is an introduction, and on Monday we are going to talk about crystal field theory, which is a topic that most people have not seen in high school. At least that has been true in past years./nIt looks a little bit scary sometimes to people because it looks different./nIt is not something that seems familiar. It is actually not that hard, but I strongly recommend coming and being present in lecture and working the problems. That is how you learn how to do this part./nAnd this often plays, this particular unit, on transition metals is a dominant theme on the final exam, so you will see it there./nIf people could settle down a little, I am even having trouble hearing myself./nThat is going to be Monday. And then on Wednesday we are going to talk about shapes of coordination complexes, and this fits in nicely with the transition metals. We are going to talk about vesper theory./nAnd then the following Monday we continue on and talk about how one can use crystal field theory to think about what colors molecules with transition metals have./nThese units, it is a little bit of a strange order this year because of the Thanksgiving holidays, but that is what is coming up./nAgain, this is a unit that most people have not seen before in high school. And, as I mentioned before, problem set nine, which is handed out today, you will see that it is really quite short. It just covers today's lecture and Monday's lecture./nIt just has a few problems on it, so it should just kind of keep you from falling behind with the material./nBut not take too much time so you will be able to get that done before Thanksgiving. All right./nToday there are a lot of topics. That is what happens when you introduce a new area. You need to go through a bunch of different kinds of definitions. And so today we are going to talk about coordination complexes./nAnd we are going to talk about coordination complex notation, coordination number, structures, just briefly of coordination complexes./nAgain, on Wednesday, we are going to be coming back next Wednesday and talking much more about these structures, but we will give a little introduction today./nWe are going to introduce something called the chelate effect and talk about isomers. And also, since we are talking about transition metals, we are going to start talking about d-electrons and d-orbitals./nThis is, again, all in Chapter 16 in your book./nHere is that relevant part of the periodic table that we will be talking about in this unit. These are what are called transition metals. And this is a part of the periodic table that is near and dear to the hearts of many faculty members in the Chemistry Department, including my own./nWe study proteins that use these metals to catalyze reactions./nAnd so this color scheme is actually out of a book that was published by Steve Lippard who is a faculty member in the Chemistry Department./nI will mention him again later in the day. He does a lot of work on the area called Metals in Biology, these transition metals that play a role in biological systems. And this book was also with Jeremy Berg who is the current head of the general medicine component of the National Institutes of Health./nAnd so he is somebody else who has thought a lot about metals in biology./nIn his case, in particular, zinc. Why are metals important in biological systems? Well, proteins, and I have talked some about proteins already, are composed of amino acids. You have carbon, nitrogen, oxygen, sulfur./nAnd there is quite a bit of chemistry that these particular elements can do, that amino acids can do, but there are some pretty challenging chemical reactions./nWe talked in the unit on chemical equilibrium about a challenging reaction, splitting nitrogen./nAnd I told you about the Haber-Bosch and how we need to do this to get fertilizers. And so it is an important industrial process that an enzyme can also do very efficiently without the high pressures and temperatures that human beings seem to need to get this reaction to go./nThat is one example of a protein that uses metals to do the chemistry./nAnd the active site of that enzyme is full of different kinds of metals. Up here says global cycling of nitrogen, hydrogen and carbon./nAnd so often some of the most fundamental reactions are also very challenging reactions and require metals to make them go. Radical chemistry reactions, making various products like vitamins or deoxy nucleotides, the building block of DNA./nAnd we have some DNA here with us today in class./nA lot of these reactions require metals. In this orange color here, those are metals that are found in biological systems, but there are also some metals in gray. And some have both orange and gray./nThere are other metals that are used as drugs or to probe biological systems./nThere is a whole area of sort of metals in medicine where people are using, say, looking at the use of gold for treating arthritis./nThere is a lot of relevance of these transition metals or d-block metals to medicine. Let's talk about the abilities of these metals to form complexes, and this is important for their ability to function in all systems, including the human body./nA key feature is that they can form complexes, and let's look at why./nAnd so these complexes then, of the metal and their ligands, are known as these coordination complexes. You have a positive metal ion, and it can attract electron density. And that is usually in the form of a lone pair of electrons from another atom./nAnd so when you bring together this metal and these other atoms and their lone pairs then you form this kind of complex./nAnd so if we look at the different roles that things play, you have a donor atom./nAnd they are called ligands. You often talk about a ligand being coordinated to a metal. And we mentioned Lewis acids and Lewis bases when we were talking about acid base, and we didn't really do much with them when we were talking about acid base./nAnd I said this would come back later./nLewis acids and Lewis bases are very relevant in terms of talking about how coordination complexes are formed. A Lewis base is going to donate lone pair electrons. And here are some examples. Again, these are referred to as ligands./nAnd so they would be acting as Lewis bases typically by donating their lone pair. Here are some of the ones that you are going to see in this unit./nAnd some of these will become familiar to you as we go through the next few lectures./nThey all have these lone pairs. And the metal is very happy to see ligands coming, and they will coordinate and form these coordination complexes. Then you need an acceptor atom, and that is the transition metal./nThat is going to act as the Lewis acid, and it is going to accept that lone pair that is being donated by the ligand./nLet's look at some example of some of these metals. And we saw some of these on the Periodic Table./nThese are all these d-block or transition metals. Then the coordination complex is the metal surrounded by the ligands or the Lewis acids surrounded by the Lewis bases. Here is a picture./nWe have cobalt in the center here, and then there are six NH3 groups which are acting as the ligands, and their lone pairs are pointing toward the metal./nAnd so, again, the coordination complex, all is meant by that is the metal surrounded by the ligands. And, in this case, there is a little bracket on the side with a plus three charge. And, for this particular complex, you might have like three chloride ions hanging around./nThree negative charged ions to kind of counterbalance the fact that this coordination complex has a plus three overall charge./nAnd so the charge would be distributed, but it is indicated that this whole complex has this charge by that little bracket and the plus three. Again here cobalt is acting as the Lewis acid. It is accepting the electrons that are being donated by the ligands./nIt is the acceptor atom./nAnd the NA3 groups are acting as the Lewis bases. They are the donor atoms, so they are donating a lone pair of electrons. Now we can introduce a term called coordination number, CN, and that is pretty simple./nIt is just a number of ligands that are bonded to the metal./nAnd so here there are six. The coordination number is six. That would be written as CN=6. And you have six ligands comprising the primary coordination sphere./nWe also have three chlorines hanging around, but they are not coordinated to the metal so they don't get counted in this. You are only interested in the ligands that are actually coordinated in the complex./nCN numbers typically range from 2 to 12, but 6 is the most common./nYou are going to be seeing a lot of different coordination complexes like this. Let me just show you how that would be written. When you are doing a problem in your book, it is mostly likely not going to be drawn./nYou are going to see notation for this instead. And so this particular one would be indicated, you have the cobalt and then you have your NH3 groups. There are six of them, so that is indicated there./nThen you have a bracket around the outside with a plus three, so that indicates that the charge of everything inside equals plus three./nYou also might see this where outside the bracket you have three chloride ions. And so you should be able to think backwards to this and say, well, if we have three things with a negative one charge./nAnd you see why the oxidation-reduction unit was important before this./nBecause we talked about oxidation numbers and all the rules for assigning oxidation numbers. You are going to need that knowledge to do this unit as well./nYou should know that this is going to be minus one for each chloride, so that would three, which would mean that if this is minus three, this bracket would be plus three, for this one down here is written to be neutral./nYou will be seeing this and need to back calculate./nThen you are also going to need to calculate what the oxidation numbers of the things within the bracket are as well. And we will talk more about that a little later in the lecture. Again, the NH3 within the brackets means it is bound to the cobalt./nThat would be this picture here. There are six of them so you would draw them as six around. And let me just mention a little bit about this particular configuration./nHere, where there is sort of a thick line coming out, indicates that it is coming toward you./nHere in this case you would have two atoms coming toward you. The dashed lines here are showing that there are two atoms back behind the plane. And then when you have the straight line, that is in the plane./nThis picture here is showing you this type of geometry, which is called what?/nOctahedral geometry, right./nLet's talk about octahedral geometry and the other geometries that you are going to see most often in this particular unit./nAgain, I am going to do much more on this on Wednesday, but this is just a very brief introduction to the structures that you will see the most in this particular unit./nFor a coordination number of six, you are just going to see one type of geometry, which is this octahedral geometry, which I just showed you./nFor coordination number of five, you will see two different types of geometry in this unit. And let's take a look at this. The first one over here, the trigonal bipyramidal./nHere is an example of that./nAnd I will try to have this as it is. Three ligands are in the plane. You have up and down and out here. One ligand is pointing out toward you with the thick line and another ligand is going back with the dashed line here./nThat is one there. We talked about the geometry./nAnd I forgot to mention that those are 90 degree angles, which I think is pretty easy to see when you look at it, that you would have 90 degrees in between those bonds./nAnd what about for this one? What do we have going for the red? What kind of angle is that? 120. And what about from the red to the black? 90. Those are the angles we would see and hear. And this is going to become important because we are going to be talking about d-orbitals./nAnd where the d-orbitals are for the metal, which is in the center of the coordination complex./nOne of the issues we will be discussing is where are the atoms with respect to those d-orbitals? Geometry becomes important at this point. The other type of coordination complex you might see, here the metal is gray in the middle, and so then we have our ligands bonded to it./nIf you have five ligands you also might see this geometry here, so this square pyramidal geometry./nAnd what are the angles between all of the atoms here? 90. Now, in the case where you have four ligands bound to your central metal there are two possibilities for the geometry that you will see./nOne is square planner. In this case, it is drawn with two ligands coming out towards you and two ligands going behind. And what would be the geometry between these guys?/n90 as well. And so this is one of the easy ones to remember because it is a square and it is planar./nThat is an easy one to remember. The other that you might see when you have four ligands is this guy tetrahedral. And what are the angles here? 109.5./nAnd then if you have three ligands, pretty much this is all you are going to see in this unit, and so we have the trigonal planar./nAgain, it looks trigonal and it is also quite planar. And what are our angles here? 120. And linear is going to give you what angle? 180. You will see this one as well./nAnd this model is a little old so it is not really so linear anymore but it is still supposed to be./nWe are going to bring back, I am going to bring these models to class for a lot of the next lectures because thinking about the structure is going to be very important in thinking about the properties of transition metals./nAll right. What else can transition metals do? What are some of the other things?/nWhen you talk about transition metals and coordination complexes, you need to talk about the chelate effect. We have seen so far ligands binding at one site, but some ligands can bind a metal at more than one site./nAnd so then you start talking about chelates. First a ligand that can bind a metal at one site is called unidentate or monodentate. And the dent comes from dentist or tooth./nThis would be one. If you have a ligand that can bind at more than one point of attachment then it is called a chelating ligand./nAnd the coordination complex that it forms is called a chelate. And so this is Greek for claws. It seems like claws are coming down and binding to the metal. Anything above one would be considered a chelate if it has two points of attachment or three or four./nBidendate means two points./nAnd this nomenclature, not too challenging. People are usually pretty good at guessing what they mean. Tridentate would have how many points of attachment? Three. Tetradentate? Four. Hexadendate? Six./nThere are some people, this was a question, I think, hexadentate on a final one year people got it wrong. Don't get something like that wrong./nSave the things you are going to get wrong for the really hard part./nThis is really not so difficult. That should be some bonus points there for you. If you remember the dentate, and this term actually gets used quite a bit, you will be all set. What is the chelate effect? Chelates are ligands that can bind at more than one point of attachment./nWhat is the chelate effect? Well, it turns out that metal chelates are very, very stable./nAnd so this is a thermodynamic property that they have. And it is believed, or the rationale for the stability which is observed, is that it is an entropic factor that is going on./nThe binding of that chelate at more than one point of attachment is entropically favorable because you are going to be releasing waters as you do this. We are going to come back, and I am going to explain this using an example in a few minutes./nFirst let me give you some examples of chelates and then I will show you how this chelate effect could work./nThere are examples of chelates in biological systems. And it may not be a surprise to some of you that one of the chelates is vitamin B12. And that would be one that I would pick to mention in particular./nHere the chelate, this is all one ligand. This whole bottom part is attached. It is one ligand./nWe have this called the corn ring. You have cobalt in the middle. And it is coordinated by this planar part./nAnd you have four nitrogen ligands coordinating. It is one ring system here called the corn ring that is attaching at four points of attachment. You also have an upper ligand here, and this is an adenosyl moiety./nAnd we have a lower ligand. And we talked about this before with buffers effects./nBecause, if this lower ligand is detached on the protein, it has to fit into a binding site. And genetic defects that block that impair the activity of the enzyme./nThat was mentioned before. This actually has six different ligands and will classify as a chelate or six different points of attachment and classifies as a chelate. Let me just sort of spin that around so you can kind of get a sense of the geometry here./nAgain, here is the cobalt in the middle, the upper ligand and the lower ligand./nAnd if we move that around, see if you can get a sense of the sort of geometry around that cobalt. You have four in a plane here, one on top and one on bottom, so it is sort of an octahedral type coordination of that metal in the center./nI thought I would just mention the person who determined the structure of vitamin B12./nThat is Dorothy Hodgkin. For her work on this, and on penicillin, she received a Nobel Prize, because at the time she was studying the structure it was really considered very large for the technique that she used, which is x-ray crystallography./nAnd people didn't think she would be able to determine where all those atoms were located in something that was that big using crystallography. Since then we have gone on to solve much larger structures./nBut she really opened the doors for people thinking about this technique as being important in biological systems./nShe had a number of graduate students who went on to do some pretty spectacular things and really helped to establish the field of crystallography. And I will just mention one of her graduate students who was not as successful as a crystallographer./nThis is what I do. And not everybody can do x-ray crystallography. It is a bit challenging./nAnd some people had to sort of find something else to do with their lives. One of her students didn't quite cut it as a crystallographer, but the good news is she was still able to find a job./nAnd it is actually kind of interesting that Dorothy Hodgkin was a real socialist. She was very involved in left-wing issues, but she stayed friend with Margaret Thatcher throughout the years. And it would have been very interesting to be in a living room with the two of them hearing them talk about politics./nMaybe they would talk about crystallography, I don't really know, if they were together./nThat might have been safer. That is one example of a chelate. Here is another example of a chelate, and this is called EDTA. And so it has many points for attachment to a metal. We have a lone pair up here on this atom, so we have a donor here./nWe have another lone pair, another donor site here. Another lone pair, another donor site here./nAlso over here and here and here. This ligand can attach to something in six places. Here, again, is the free molecule./nAnd here is what it looks like when it is attached to the metal. See this oxygen here, over here, and then if you follow it down over here this nitrogen is also coordinated. If we go over here then this oxygen is coordinated./nIf we come from this nitrogen over to the other nitrogen it is over here./nGo down to this oxygen, this oxygen is coordinated down here, and then in purple we have the last coordination. What is the geometry around that metal site? You have these six ligands, so what is that going to be?/nIt is going to be an octahedral./nYou should have angles around 90. And it is also hexadentate, so you have a ligand binding with six points of attachment. It would also have a CN number then of six. Why the chelate effect? Why are complexes like that so stable? The chelate effect could be viewed in this slide./nBefore you have the EDTA bound we have six water molecules around it./nThen we bind one molecule of EDTA. It binds in the hexadentate complex releasing all six waters. Where is entropy greater? When you have one free thing or when you have six free things? Six./nThis is entropically favorable. You are creating disorder./nYou are binding one thing and releasing six things. That is why these complexes are really quite stable. We are getting back to entropy now./nWe have been talking about delta G. We had delta H. Now we have delta S. That is the chelate effect. And we talk about it due to this entropic effect accompanied by the release of non-chelating ligands, usually water, from the coordination sphere./nIn this case, one ligand binding displacing six things is favorable./nThe chelate effect leads to a lot of practical uses. Here some of the uses would be in medicine. If you discover over Thanksgiving that you one and a half year old niece or nephew has discovered paint on the windowsill as peeling and is starting eating the paint off of the windowsill, you might want to remember when that was painted last and if there was lead in that paint and rush the child to the emergency room./nAnd when you get there, just in case the people there, you know, on Thanksgiving it is always dangerous to be in an emergency room./nYou might say give this kid EDTA. That is really what is used. And that will chelate then the lead out of the system before it can bind to things and do a lot of damage./nThis is one of the chelators that is used./nIt is a chelate effect. It works pretty well if someone has been exposed to something like lead, which is pretty toxic and you want to get it out of their system. That is one use. It is also in food./nIt is a food additive. Sometimes it is fun after you take a couple chemistry courses to actually start reading the labels of the food that you are consuming and some of those words will actually start making sense./nOne thing that you will often see is EDTA./nAnd it will say EDTA was added for freshness. One thing that EDTA can do. Microorganisms, bacteria, other things that you would rather not have consuming your food at the time that you also want to consume it need metals for life, as we all do./nIf you add EDTA, it is a good way of preventing other kinds of unwanted microorganisms from growing on the food./nBecause it will chelate all the metals and the organisms cannot live. Added for freshness is a more polite way of saying that this food is less likely to be contaminated with lots of bacteria growing on top of it./nAnother thing that it can be used for is bathtub cleaning or other kinds, but bathtubs and showers in particular. There is calcium in tub scum./nAnd most kinds of cleaners have something that chelates out that calcium which allows you to clean./nAgain, over Thanksgiving, one thing that happens to a lot of people when they go home is that their parents say help us clean to get ready for everyone coming. And so you might be using EDTA then as well./nI thought I would tell you a little story about how knowledge from freshman chemistry can make somebody money./nThis is a really kind of fun story. I don't know if it was around Thanksgiving, it might have been./nOne day Robert Black had a life altering experience. His wife said to him could you clean the tub? Apparently Robert Black had never cleaned the tub before that time, and so he went to clean it. Apparently, maybe it had not been cleaned in a while./nAnd he discovered that it is really hard./nYou have to scrub a lot to get the tub clean. He thought wouldn't it be nice if one never had to clean a tub? He came up with this idea. And here is one of the products that came out of this, Sparkling Shower, available at Leveret's./nAnd what he did was he put the same ingredients, including EDTA, in that you have in most cleaners, but he had sort of a novel pitch./nEvery time you take a shower, just squirt the shower or the tub, and then you never really have to clean it./nIn this patent he had a surfactant chelating agent, EDTA and an alcohol. You had a surfactant to sort of break up the water tension, the chelating agent to sequester out the salts in the scum and alcohol to dissolve the more oily ingredients./nAnd so those together kind of got everything loose and ready to go./nAnd the surfactant kind of helped it run off. If you squirt your shower every time you shower, you will never really have to clean it./nHe used all ingredients that were know before but packaged it slightly differently. Again, EDTA is one of the ingredients in these products. And so sales of these particular products have reached $70 million every year./nI think that the wife is probably very happy that one day she insisted that he clean the tub./nBecause this turned out to be a pretty profitable exercise. It is also an example of how some very simple chemistry can yield you quite a bit of money./nAll right. Tub cleaner. There was one more that I thought of. Anybody a fan of vampire movies?/nNot really all that many. It is a little surprising. I am not sure you are all telling the truth./nIf any of you happened to see Blade, you might be aware of another use for EDTA. If you remember, you may not have noticed what they were mentioning at the time, but they came up with a great way of killing vampires./nThere were these darts filled with EDTA./nI think the kind of general premise that you can rationalize this Hollywood special feature is that vampires drink blood, blood has iron, and if they are pretty much all blood, if you give them something that chelates out the iron it could cause some kind of harmful effect./nIn the movie they exploded and sort of disintegrated completely. If you take your niece or nephew to the emergency room for eating lead paint this will not happen./nI am quite sure. But it was really very spectacular./nHere they are, the little darts filled with EDTA. Again, more knowledge that could be useful on some occasion, perhaps, I am not really sure. I have a feeling that most of you will find the top three more useful, but here is the complete list that I have been able to come up with for the uses of chelators./nAll right./nThose are some coordination complexes, either single ligands bound around a central metal or chelating ligands bound around a central metal. There are other coordination complexes which have important features./nAnd one thing that we will talk about is isomers. Geometric isomers can have very different properties to them. It seems like it is pretty much almost the exact same compound./nBut the arrangement of the atoms is slightly different and that can yield two very different properties for those particular compounds./nLet's look at one example. This is platinum, and it has two NH3 ligands and two chloride ligands. And so that is in the bracket. Those four different ligands are coordinated centrally to the platinum./nIt would have a CN number of four./nAnd the most common shape that you would see for this would be this one here, this square planar shape. Depending on how you arrange the four different ligands two NH3 and two Cl, you get a very different result./nWe have, in this arrangement, a compound called cisplatinum./nYou have a cis arrangement where the chlorides are on one side and the ammonia groups are on the other side, or you can have transplatinum where the ligands are across from each other. Here are these two different molecules./nHere they are cis with respect to each other and here they are trans with respect to each other./nOne is a very potent drug that is used in treatment of cancer. The other has no known function. They are composed of identical atoms./nWhy is this different? Why are there different functions with the different geometries? Well, as it turns out, it is real important to have the atoms on the same side./nAs shown in this slide, I said I would mention Steve Lippard again, and here I am, this has been a subject of research in Steve Lippard's laboratory here in the Chemistry Department at MIT for a number of years./nIf you have a cancer cell and you want to kill the cancer cell, one thing that you can do is give it cisplantinum. The cisplatinum comes and binds to DNA. Here is a little cartoon of DNA here./nThere is the actual model of DNA structure down here in the front of the room./nWhen it binds then it can inhibit transcription. And so this can then lead to cell death. And so Steve has been really looking at this issue trying to redesign cisplatinum to see if you get a better drug, looking at other factors which might have an additive effect./nThis particular drug is used often for prostate cancer and has really been quite effective./nWhy does it matter if they are on the same side? Well, if you look at this complex and the other, you see that only two of these ligands are left./nTwo of the ligands are gone. At those sites it is actually coordinated to the DNA itself. That is why it is important that they are on the same side. You want to release two chlorides and have it bind to DNA this way./nIf it is on the other side, it really doesn't work as well. If we go over and look at the DNA molecule for a minute, you see that if you want to have two free sites to coordinate two DNA molecules./nOf course, this is the totally wrong scale./nYou want these on the same side. If I made it on the right scale it would not fit in the room, really. This is more or less, you can imagine. If you are going down here and you want to release just the chloride ligands, well, you cannot get the chloride ligands to be on the right side for this./nMolecules that have different shapes may have different properties./nAnd if you are in an environment you are looking to react with a particular molecule. I will show you another kind of view of DNA./nDNA is a really beautiful molecule. Here you have something you can carry around and show your friends, but also there is a smaller model of DNA. This is actually an artist in Michigan who will carve out within these blocks any atomic structure that you have./nAnd you can show the sort of bases in the detail./nThat is what DNA looks like sort of from the top and then from the side. It is just a really beautiful molecule. This is a nice example. Steve Lippard often takes a lot of UROP students, so it might be something, if you are interested in this area, that you might want to look at his Web page and see if that type of research interests you./nI lost my pointer./nThe cisplantinum has the effect because you need to release chlorides on the same side so that it can bind DNA over here. This cannot bind DNA because you don't have the chlorides together so they cannot be released and you won't interact with the DNA./nAgain, these different isomers have really, really different properties. Here we look at optical isomers or enantiomers./nAnd introduce the idea of a mirror plane here. When you have two different isomers, and if they are not superimposable, that means there is a mirror plane and that they are chiral./nAnd chiral molecules can have vastly different properties, again, if they are in a chiral environment such as the human body. Let's just look at these two./nSee that there is a mirror plane. Here we have the green on top, the NH3 on top, chloride in gray here on the bottom, and then on the side we have the oxygens, the black atoms here, more chlorides here, on the other side there are also oxygens pointing out in front, another NH3 group sort of putting in back./nAnd so they are very, very similar./nBut if you try to superimpose them, you will realize that it doesn't match up. If you try again other ways, put another green over here, no, that is wrong, too. And you can spend maybe not very long convincing yourself that, in fact, these are non-superimposable./nThey are mirror images so they are chiral compounds. And the idea of chiral compounds, we will come back, and you will hear much more about that if you take organic chemistry 5.12 or 5.13./nNow we have to get to some business that we need to think more about the properties of these elements./nAnd that is we need to count. A lot of counting in these units, oxidation-reduction, and also in this unit counting skills. We need to figure out how many d-electrons are involved in coordination complexes to be able to rationalize their properties./nThis is pretty straightforward./nOne can figure this out, what the d-electron count is if you know the group number from the periodic table and the oxidation number of the metal. We know how to calculate oxidation numbers at this point and the Periodic Table in your book./nLet's just take a look at a couple of examples and count some d-electrons. Suppose we have a coordination complex./nIt has cobalt in it, it has six ammonia ligands, and it has a charge of plus three for the overall complex./nThen we need to figure out what the oxidation number of cobalt is first. What is the overall charge on NH3? That is zero. This is a neutral molecule. And if you don't know that, this is something you will memorize as you go through this unit./nWhat does that leave you for cobalt? Plus three, because it all has to add up, just as it has before./nAnd so we figure out the oxidation number first. And then second we would figure out the d-count. We need to look at the Periodic Table and see where it is. Here we have cobalt in Group 9./nWe have Group #9 minus the oxidation number, which is three that we just calculated, and we are left with six./nAnd you will often see this written as d6 for the d-count number of electrons. One year when I had this, I think the first year, I actually had, I think, nine minus three equals five in the handout./nAnd that was rather embarrassing./nOh, wait. I want to do one more example first, or two more examples first. Let's try just a couple of more. All right. There are some that will become very familiar to you as you do these. Let's look at nickel with CO ligands, four of them, written like this./nThere is nothing up here./nThat means that the overall charge has to equal zero. CO, that is one that you will come across, and its overall charge is also going to be zero. It is neutral as well. And so what does that leave for nickel? Zero./nSometimes not very exciting. All right./nIf your oxidation number is zero then we want to look up and figure out what the d-count is. We have to see where nickel is. It is in Group 10, so we have ten minus zero./nIt is a d10 system. Let's just try one more./nCobalt, again, two molecules of water and one of NH3 and three of chloride to the minus one. Let's put that up a little bit. All right. Overall, this has to be minus one./nWhat is the charge on chloride going to be? It will be minus one./nThere will be three of them. NH3 was what again? Zero. Do you want to guess what water is going to be? Zero. Two zeros. What does that leave you for cobalt? Plus two. All right. Cobalt we looked up before and it was in 9./nOver here we have plus two./nAnd we have Group 9 minus two is seven, and so that is a d7 system. That is how you will do those problems. And you will need that on Monday when we start talking about crystal field theory./nIn the last couple of minutes, I want to remind you of the shapes of d-orbitals. This is leading into what we knew next./nAgain, we are thinking about the shapes of the d-orbitals and the metals, and then we are thinking about the shapes of the coordination complexes./nAnd we are trying to figure out whether the ligands are going to be near the d-orbitals or away. And depending on how the ligands are arranged with respect to the d-orbitals changes the properties of the coordination complex./nAll right. There are five d-orbitals. And you should be able to draw them./nAgain, the bar for successfully drawing the d-orbitals is you have to do as well as I do when I try to draw the d-orbitals./nAgain, it is a low bar for drawing. You just kind of need to get some of the general features right. If we look at dz2 it has a maximum amplitude along the z-axis. And in this course we will always try to keep the axes more or less the same to help you out./nAnd they will be labeled so you know where the z-axis is. Here is the z-axis and y is along the plane here./nAnd x is coming out toward you. There is x. And then you have a doughnut in the xy plane./ndx2 minus y2 has maximum amplitudes along x and y. It doesn't have anything along z. The orbitals are along y in the plane of the board and along x coming out toward you and also going behind the screen./nThen we have three more./nWe have dyz and it has its amplitudes 45 degrees off the y and the z-axis. Here is z up and down and y across the plane of the board. And so these orbitals are drawn to be 45 degrees off those axes./ndxy is 45 degrees to the X and the Z, so these are coming out towards you and going behind and also going up./nAnd then dxy is 45 degrees off of y and x. And you will need to know how to draw these and need to know the shapes and to be able to think about them in 3d so that you can imagine where ligands are and how those ligands are affected by these d-orbitals./nHave a great weekend.
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Added: April 16, 2009, 11:38 pm
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Chemical Science - Kinetics (cont.) - Lecture 33
Principles of Chemical Science/nVideo Lectures - Lecture 33/nTopics covered: /nKinetics (cont.)/nInstructor: /nProf. Catherine Drennan/nTranscript - Lecture 33/nWe are getting close./nWe are in the last unit. We are doing kinetics. And we are going to talk about reaction mechanisms today. We had a little bit of an introduction to that already last lecture setting up to do this./nToday we are going to run through and talk about how you come up with reaction mechanism, how you evaluate a proposed mechanism in terms of the experimental data. This is, again, in Chapter 13. All right./nSuppose you have a particular reaction. You know that 2NO gas plus O2 gas go to 2NO2 gas. And someone has come up with some experimental data for you. And they figured out that the rate law for this reaction, there is observed rate constant, so K observed./nAnd in terms of the NO concentration, it is second order. And in terms of O2, it is first order. And so now you can come up with a reaction mechanism that fits that experimental data. And, as I mentioned last time, you never know for sure that your reaction mechanism is correct./nAll you can do is be consistent with the data. You can prove reaction mechanism is wrong. You can come up with a mechanism and show that that cannot possibly be the mechanism given the experimental data./nYou can prove things wrong, but you can never be sure that you have it right. The best you can do is consistent, but you can prove a lot of things wrong. And if you have a lot of mechanisms and you prove a lot of them wrong and there is only one that seems to hold then there is a good chance that probably is correct, but you never 100% because someone might do another experiment and discover something else./nBut, in the problem set, you will be saying, yes, this mechanism is consistent, or you will be writing a consistent mechanism, or you will be saying that mechanism cannot possibly be true. All right./nLet's look at that. First let's just consider what the overall order of this reaction is. What is the overall order if this is the rate law? Three. I am just running through these. We will practice some of these because these are good couple point questions on the final, and so you don't want to miss these easy points./nAll right. If the overall mechanism is three, the overall order is three, three things coming together at once to react. Is it likely that is it one step? No, it is not. What is it called if you have three things coming together in terms of molecularity? Termolecular./nAnd those are rare, as we talked about last time. As I said, if you are getting together three people, it is usually not true that they all arrive sort of ready to go at the same exact moment. Usually one person is late, if not more./nGetting three things to react together at the same time is challenging, and so those are rare. So, this probably has more than one step. Then we have to try to divide up this reaction into steps and come up with a mechanism, a two-step mechanism we will try for this./nLet me leave that up. All right. Let's draw a potential mechanism on the board. And then we will try to figure out, for that mechanism, what the rate law would be if that was the correct mechanism and see if it is consistent with the experimental data./nAgain, what was the experimental rate law is K observed. That is the observed rate constant. Second order in NO and first order in O2. We can write the reaction in two steps, and each step is an elementary reaction so it occurs exactly as written./nIn the first step, we will have two NOs coming together. And the forward rate constant will be k1 and the reverse rate constant will be k-1. And it will be forming an intermediate N2O2. In the second step of the reaction we will have oxygen come in and interact with the intermediate, N2O2, and form, with a rate constant k2, two NO2s./nAnd that will give us our overall reaction. Now we need to write rate laws for each step. And since each step is an elementary reaction, we can write the rate law using the stoichiometry of that step as written./nThe rate law for the forward reaction would be what? What do I put first? k1. And then? NO squared. What about the rate for the reverse reaction? k-1. Times what? N2O2. OK. Let's look at each of these./nFor the forward direction, what is the order? The overall order is two. And so in terms of molecularity what is it? Bimolecular. For the reverse reaction the order is what? One. And what is that called? Unimolecular./nOK. Let's look at Step 2. The rate for step two is what? k2. Times? It would be O2 and N2O2, the concentration of each. Again, we can write them exactly as written. Now we consider the rate of N2O2 formation, so the overall rate of NO2 formation from these two steps./nAnd we can just use second step where NO2 is being formed. And so we can write this is equal to that rate law times two. Let me rewrite the rate law. And the two comes from the fact that two NO2s are being formed./nBecause two NO2s are being formed, this concentration is increasing at a different rate than the concentrations of these are decreasing. The book is inconsistent with whether it puts a two all the time, so I am not going to take off if people forget to put the two in this./nBecause the book should be more consistent with that. But, if you do see a two, it will mean that you are forming two of the products in the reaction. All right. Now we have written a rate. We are not equal to our experimental rate here because there is an intermediate./nThis is an intermediate. And an intermediate cannot appear in the overall rate law for something. You are going to be spending a lot of time getting rid of intermediates. Let's now leave this here and try to get rid of our intermediate./nI am having trouble with papers today. Now let's solve for the intermediate. And we want to solve for this intermediate in terms of things that are not intermediates. We want to solve for it in terms of reactants or products or rate constants./nWe need to consider then how this intermediate N2O2 is being formed and how this intermediate is being consumed or decomposed. We want to figure out the net formation of the intermediate. And the net formation of the intermediate is going to equal the rate at which it is formed./nIt is being formed over here in the forward direction of the first step, so we will write that down. We consider how it is being formed. It is being formed with the rate k1 times the concentration of NO squared, so that is just the rate law for the forward direction of the first step./nIt is also being decomposed in the first step in the reverse direction. It is being decomposed with the rate of k-1 times the concentration of the intermediate. That is just our rate law for this reverse step of the first reaction./nAnd it is also being consumed. It is also being consumed in the second step with a rate law of k2 times the concentration of the intermediate times the concentration of O2. That is just the rate law that we wrote for this second step./nIt is being formed, decomposed and consumed. The net formation equals the rate at which it is formed minus the rate at which it decomposes minus the rate at which it gets consumed. And now we can use something called the steady-state approximation./nIn the steady-state approximation, all it is saying is things are in a steady state. That means that the net formation of this intermediate is equal to zero or, if it is in a steady state that the rate at which it is being formed is going to be equal to the rate at which it is being consumed, either by the reverse Step 1 or Step 2./nThis rate is equal to this rate. That means it is steady, so it is in the steady state. The net formation is zero. The rate at which it is being formed equals the two rates at which it is being consumed or decomposed, the rate at which it is decaying are equal to each other./nThat is the steady-state approximation, and it is a very good approximation. And so you can use this for all of the mechanisms that you will be writing. That allows us to set it equal to zero, or we can rearrange these terms so that these two are on one side and the rate at which it is formed is on the other side./nLet's do that. And I am going to take these two and put them together and also pull out the term for the concentration of the intermediate because I want to solve for that term. I am going to try to pull it out and solve for it./nIf we rearrange this then, we get that the concentration of our intermediate, you pull it out from those terms, so that leaves k-1 plus the other term k2 times O2. Now I have just pulled this out from these two, and I have moved these to the same side of the equation./nAnd so that is going to be equal to k1 times NO squared. This is the rate at which you are decaying your intermediate and this is the rate at which you are forming that intermediate. Now I can continue to solve for this./nThe concentration of the intermediate is going to be equal to k1[NO]2 over k-1 plus k2 times the concentration of O2. Here is what the intermediate then is equal to. And now I can take this term and go plug it back in./nAll right. We can take the concentration and plug it back into here, so that whole term that we just came up with. And we will put it back into that equation. Then, if we substitute, we get the rate of NO2 formation is going to be equal to two rate constant k1, rate constant k2 times NO squared times the concentration of O2 over k-1 plus k2 times the concentration of O2./nBut that is not consistent. We have NO squared and two O2 terms. Whereas, in our observed experimental law, there is only one term for O2. That is not consistent so something is wrong. This is not consistent so there must be fast and slow steps./nIf there are fast and slow steps then we can simplify this equation and make it like the experimental equation. Now we can go over here and think about what would happen if Step 1 was fast. It is a fast and it is a reversible step./nIt is already written that it is a reversible step, so it is reversible. And Step 2 might be slow. Now we can consider, if we have a fast first step and slow second step, what that would do to this equation and whether we could then rearrange this equation such that it would be equal to our experimental rate law./nWhen we say something is slow, we are going to introduce a new term. And that the slowest step in a reaction mechanism is called the rate determining step. And, for a rate determining step, you can assume that that step is governing the overall rate of the reaction./nIt is slow enough that the rates of the other reactions don't really matter. That is what rate determining means. And let me just give you an example of a rate determining step. After today's lecture, you will be able to do all of the problems on Problem Set 10./nThis is your last problem set in 5.111, and so it would be really great when that problem set is finished and turned in. And so all through the rest of this lecture thinking about I can go out right after class and finish that problem set because you are very excited./nAnd by the time it gets close to the end, you see the end of the handout coming up. Your books are sort of packed, you are ready to go, and then the minute it is finished you are out the door. And you are going to some place where you can sit down and finish Problem Set 10./nYou are moving really fast. You might even be running. You may be even jumping over people in the infinite corridor to get to the library as fast as you can. You get to the library and you look for a table, but everyone else from 5.111 is already there and all the tables are filled./nYou go to the first floor and the second floor and the third. You go to the basement and in the stakes. Every table is filled up with people doing Problem Set 10. Then you have to leave the library./nYou think Building 2, lots of small classrooms, they cannot all be in use. You run over to Building 2, but all those classrooms are filled with 5.111 students doing Problem Set 10. And so eventually you get to Building 4 and you find a free classroom./nThe minute, the second you find that free classroom your books are out of your bad. They are out. Your calculator. Your correct kind of approved calculator is right there and you are ready to go. It takes you maybe two seconds to get out [10 at 2:50?] and it takes you maybe two seconds to pull out all your books to be ready to go, but it takes you a half an hour to find the table to do that problem set./nOverall, half an hour plus four seconds is a half an hour. That was the rate determining step. In finishing Problem Set 10 was finding that table. What you can do for these as well, if we say there is a slow step, you can assume that that is rate determining./nAnd you don't have to worry about how fast those other steps are. It just depends on the rate of the slow step. That is governing the overall rate at which the reaction or the process happens. That is rate determining step./nWe can use that now. And we can use that and go back and simplify the expression that we solved for the intermediate. Here we were solving for this expression, and we weren't considering anything about fast and slow steps at this point./nAnd so this is the equation that you would get if there were no fast and slow steps. And then we plugged it into this and found that it was not consistent. Now let's take a step back and go back here and think about what this expression would be if the first step was fast and the second step was slow./nHere, again, I have just put it up on the screen what we had on the board. We have this first step is fast and reversible and the second step is slow. Really, what we are talking about here is we are comparing rates for the decomposition of the intermediate, that is the reverse direction of step one, with the rate for the consumption of the intermediate./nWhich of these is faster? Well, decomposition is faster because this is a fast step. And the consumption in Step 2 is slow. What does that mean? Well, if we write it down then if the rate of the decomposition, this rate, the rate of the reverse reaction is faster then this k-1 is a bigger number than k2./nThe rate of consumption here is slow. We are going to consider the relative size of these numbers and how they fit in this term, and you can drop things out by assuming that one thing is a lot bigger than the other thing./nIf we do this then we are saying, again, this is the same thing that was on the last slide, that the rate of the decomposition is faster than the rate of consumption. i.e., Step 2 is slow, Step 1 is fast./nWhat you are saying is that k-1 is a bigger number. That rate constant is a bigger value than the k2 term. And, in the bottom of the equation here, in the bottom of this equation for the intermediate, we have a term k-1 and we have the term k2 times the concentration of O2./nAnd now, if you are saying this about the different rates of this reaction, if you are saying this is fast and that is slow, this is a big number, that is a smaller number. If you have on the bottom here a really big number plus a much smaller number in comparison, this smaller number is not adding much to the bigger number./nAnd we can just cancel that term out. We can just say it is so much smaller than the other, we are not going to worry about it, and we are going to get rid of it. Now our expression would look like this./nThe concentration of the intermediate is k1 times NO squared over k-1. We can rearrange this and think about does this look familiar to any of you in terms of what we have seen before. We can rearrange it to look like this./nAnd what is this term? What is this all equal to? K, the equilibrium constant. It is equal to k1. This is just an equilibrium expression. And, if you go back and look over here, you will see that if you are going to write the equilibrium expression for Step 1, products over reactants, that is what you would get./nAnd also, we saw last time, that you can express equilibrium constants in terms of rate constants. And the equilibrium constant for a reaction is going to be equal to the rate constant for the forward reaction over the rate constant for the reverse reaction./nAnd at equilibrium those rates are equal to each other. So this is just an equilibrium expression. When we assumed Step 2 was slow and Step 1 was fast, we can now solve for the intermediate in terms of an equilibrium expression./nLet's consider why this is true. OK. When you have a fast first step followed by the slow step, the first step is pretty much in equilibrium. The first reversible step is in equilibrium. And you can kind of think about it in terms of this diagram./nHere you have reactants going to intermediates, and this is fast and reversible. Then the second step, the consumption of the intermediate is slow going to products. If this is really pretty slow and not much of the intermediate is being siphoned off to product, more of it in fast equilibrium is going back and forth with the reactant, then that creates sort of an equilibrium expression./nIt is not perfect. Some of it is being siphoned off, but more or less you can consider this in equilibrium. Again, if you have a fast reversible step followed by a slow step, you can assume equilibrium conditions./nThat is going to make your life easier in terms of solving these expressions, because you can go right to writing an equilibrium expression then for your intermediate. Now we can go back and substitute./nWe have this term now for solving for our intermediate. Or, you could write rate constant k1 over k-1. Or, you could write big K, the equilibrium constant k1. These are equal to the concentration of the intermediate./nWe can put this term back in here now. This was, again, what we wrote in the very beginning for the formation of NO2. And we can substitute that back in. And we can have it in terms of the rate constants or using an equilibrium constant in there./nThose would be equivalent. And all of these Ks, in terms of the experiment, are going to be grouped together as a K observed. You usually are not going to measure all the individual rate constants./nYou are just going to have some overall rate constant that you do measure, an observed rate constant. And so if you write then either of these in that form where all your Ks just get grouped into K observed then you see that that is consistent with the experimental data./nThat mechanism works. A mechanism works that has two steps, the first step being fast and reversible and the second step being slow. We don't know for sure that that is the mechanism, but it is consistent with the experimental data./nLet's look at another example now. We are going to look at three different examples. Here there is a proposed mechanism that is very similar to the one that we just looked at. Now we are going to try to do this problem in a more efficient way than we did last time, which is kind of work through it the way you will when you are doing Problem Set 10 really fast./nAll right. We have a two-step mechanism, one that is fast and reversible and a step that is slow. The first thing you want to do is write the rate laws for each of the individual steps. For this one, what is the rate law for the forward reaction? k1 times O3./nThat is for the forward reaction. For the reverse reaction we have k-1 times O2 times the concentration of this intermediate O. Let's look at the second one. Here we have the rate k2 times the intermediate O times O3./nNow the rate is going to be determined by this slow step. And we can write the equation for the formation of O2 from this last step. And we could do that whether it was the slow step or not. We did that before as well./nIf we write the formation of O2, it is being formed in the last step from O and O3. We are using this rate law. And, again, we have a two in there because, like the last example, two things are being formed./nWe have to molecules of O2 being formed. The rate of formation of O2, two times k2 times this intermediate times O3. If there is no intermediate in the expression we would be done, but there pretty much always is going to be an intermediate in the expression./nWe need to solve for that intermediate. O is an intermediate, so you need to solve for it in terms of reactants, products and rate constants. Again, here we have this step that we saw before. The first step is fast and reversible, the second step is slow, and so you can assume, approximate that the first step is in equilibrium./nAnd so you can solve for the intermediate in terms of an equilibrium expression. What is the equilibrium expression for the first step? K would be equal to? I will just put it down here. You can write it in terms of k1 over k-1, products O2 times O over reactants O3./nThat is just our equilibrium expression. We have it in terms of the rate constants and in terms of this expression for products over reactants. And now we can take this and rearrange it to solve for our intermediate./nWe would have this, we would bring the O2 down to the other side, we would bring the O3 up here, which leaves us with k1 times the concentration of ozone over k-1 times the concentration of oxygen gas./nThat was a lot simpler than what we did over here. And so you can feel free, if you are told there is a slow step and a fast step, just jump right to that and solve it in terms of the equilibrium expression./nIt will save you some time. You should come up with the same answer if you do it the way that we did it in the first step. And sometimes you will have to because you won't know about fast and slow steps./nNow we can take this term and substitute it back into the rate law for the formation of O2 gas. And so we had two k2 times this intermediate times O3. We put this term in for our intermediate and we get two times k2 times k1 ozone squared over k-1 times O2./nAnd so this would be the rate law for this particular mechanism. If that is the rate law for that mechanism, you could think about what are the orders in this and could you experimentally go back and look and say, OK, is that consistent with the mechanism? Let's take a look at what we would predict from this./nFirst we could also write this in terms of K observed and group all the K constants together for K observed, and that is times ozone to the two and O2. If we look at this, if we were going to design experiments for someone to do, what would we predict to sort of prove that that is, in fact, a correct mechanism? What is the order, based on this rate law, in terms of the concentration of O3? Two./nIf someone went into the lab and doubled the concentration of O3, what should they observe happen to the rate? They should observe four times the rate. What is the order in terms of O2? What is it? Minus one, right./nIf you doubled the concentration of O2, what should happen to the rate, keeping everything else the same? It should be half. This is the kind of thing that can be experimentally tested, whether it is consistent./nAnd what is the overall order then of this reaction? It would be one. And so if you doubled both things, what should you observe happen to the rate? Yeah, it should double. You could go back and sort of test this./nOK. Let's look at one more example. This time we are not going to make a proposal ahead of time about fast and slow steps, but we are going to work out what the overall rate law for this is without fast and slow steps and then go back and consider which step, or any, are slow./nNow we are going to go through the long way and do this problem and then think about what we can learn from it in terms of we are given an experimental rate law. And once we see how this works out, we can think about whether a step is slow or not./nFirst we start the same way, and we are going to write the rate laws for each of the individual steps. Again, they are elementary reactions. Steps in a mechanism are elementary reactions. We can write the rate law exactly as the equation is written./nWhat is the rate law for the forward direction here of Step 1? First we have k1 times NO concentration times BR2 concentration. That is the forward. And in the reverse we have rate constant k-1 times NOBR2./nFor step two, we have rate constant k2 times NO. There you go. So, you know how to do these. We have that. Now we are going to be looking for the overall formation of NOBR, and so we can write this just using the last step./nWe can use that rate expression for that last step. The formation of this. And we are putting a two in because, once again, there are two of these being formed in that last step. Again, we could just use the rate expression for the last step./nAnd there is a two in there because two are being formed. That is the rate of formation of NOBR. Now we have an intermediate. This is being formed and is being consumed, so we need to solve for the intermediate./nThere cannot be any intermediates in our overall expression, so we need to get rid of that. So, it is intermediate and we need to solve for it. Now, we cannot go and just use the equilibrium expression for the top because we don't know anything about fast and slow steps./nWe don't know that that is going to be a good assumption. Let's just write it out the long way and then come back and consider whether that is consistent with the experiment or we have to do some modifications./nWe are going to solve for this in terms of reactants and products in Ks. Now we want to consider the net formation or the change in the concentration of this intermediate. The intermediate is being formed in the first step, so we can write down the rate expression for how it is formed, which is just this, k1 times NO times BR2./nThis is how it is being formed. It is decaying in this second step, in the backwards step, so we can put that in. We have a minus sign here. Because it is going away by the rate expression of k-1 times the concentration of the intermediate and it is also being consumed in the second step./nIt is being used up in that second step, so we will put that term in here. That is minus k2 times the concentration of the intermediate times NO. It is being formed in one step and it is being consumed in two different steps./nNow we can use the steady state approximation which says that the overall net change is zero. We can set this term equal to zero. It is a steady state approximation. That is the same as saying that the rate at which an intermediate is formed is equal to the rate at which the intermediate goes away./nWe can rearrange these terms, bring these on one side of the equation and have this one on the other. Rearranging then we moved these onto one side and the formation to the other side. Now we can pull out the term for our intermediate./nTake it out here, that leaves k-1. Take it out here, that leaves k2 and NO. Again, this is the rate at which it is being formed. And now we can solve for the intermediate. When we solve for the intermediate, we take this term and divide it by that term and we get this expression./nThat is solving for the intermediate. The next step is to take that intermediate and plug it back into the initial equation that we had, the formation of NOBR, and that was two times k2 times the intermediate concentration times NO./nNow we take this whole term and put it back into this term and we get this expression here. All right. This expression now is not consistent with this experimental rate law, so there must be fast and slow steps./nNow let's consider what is going to happen if the first step is slow and the second step is fast. Again, you are considering what is happening on the bottom of this expression. You are asking how does k-1 compare to this k2NO term./nAnd so if you say that the first step is slow then k-1 is small. Second step is fast. That means k2NO is a bigger term. If this second step is fast, the rate constant for the second step is a bigger number than the rate constant for the first step./nAnd so what you can do then is you can say, OK, that is small compared to this so we drop out that term. If we drop out that term, can we cancel anything else? What else can we cancel? We can cancel k2./nWhat else? We can cancel out one of the NOs. And that leaves us with this expression here. There is the expression again, two times k1NOBR2. And that can be written as K observed. And, look at that, it is consistent with the experimental rate./nNow let's just check the other way to make sure one should be right and one shouldn't be right. That we can get the answer if we go the other way around. This is consistent with the experiment. And the overall number of this is two./nAll right. Now let's look the other way around. If the first step is fast and the second step was slow that would mean that k-1 would be a lot bigger than the k2 term. k-1 rate constant for the first step./nk2 rate constant for the second step. Fast. Bigger than this. And so then we can take this term out. Can we cancel anything else? No, you cannot cancel anything else. This is the expression. We can put that down here./nWe can group the K terms to K observed. And so this is not consistent with the experiment that you had before. That way doesn't work. You can write out the whole thing and then think about whether they are fast and slow steps./nAgain, you will be asking the question which of these terms is bigger than the other term? What is the fast step? How does k-1 compare to k2? You can make the simplification. And only one should be consistent./nOne mechanism should be consistent. The other one should be inconsistent. OK. Now you can do Problem Set 10.
Tags // Kinetics
Added: April 16, 2009, 11:43 pm
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